Count ways to increase LCS length of two strings by one
Last Updated :
23 Jul, 2025
Given two strings consisting of lowercase characters, the task is to find the number of ways to insert a character in the first string such that the length of LCS of both strings increases by one.
Examples:
Input : s1 = “abab”, s2 = “abc”
Output : 3
Explanation: LCS length of given two strings is 2. There are 3 ways of insertion in s1 to increase the LCS length by one:
s1 = “abcab”, adding character 'c' at index 2.
s1 = “abacb”, adding character 'c' at index 3.
s1 = “ababc” adding character 'c at index 4.
Input: s1 = "abc", s2 = "abc"
Output: 0
Explanation: LCS length cannot be incremented.
Input : s1 = “abcabc”, s2 = “abcd”
Output : 4
[Naive Approach] Using Brute Force Method
The idea is to simulate the insertion of every possible lowercase character (a to z) at every possible position in S1 and compute the LCS of the modified S1 with S2. If the LCS length increases by one after the insertion, we count it as a valid way. This approach checks all possible combinations of insertions and directly verifies whether the LCS length increases.
Please note that we have directly used DP Solution of LCS Problem in the below code.
C++
// C++ program to get number of ways to increase
// LCS by 1
#include <bits/stdc++.h>
using namespace std;
// Returns length of LCS for s1[0..m-1], s2[0..n-1]
int lcs(string &s1, string &s2) {
int m = s1.size();
int n = s2.size();
// Initializing a matrix of size (m+1)*(n+1)
vector<vector<int>> dp(m + 1, vector<int>(n + 1, 0));
// Building dp[m+1][n+1] in bottom-up fashion
for (int i = 1; i <= m; ++i) {
for (int j = 1; j <= n; ++j) {
if (s1[i - 1] == s2[j - 1])
dp[i][j] = dp[i - 1][j - 1] + 1;
else
dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
}
}
// dp[m][n] contains length of LCS for s1[0..m-1]
// and s2[0..n-1]
return dp[m][n];
}
// Method returns total ways to increase LCS length by 1
int waysToIncreaseLCSBy1(string s1, string s2) {
int n = s1.length(), m = s2.length();
// Find the original LCS.
int lcs1 = lcs(s1, s2);
int ans = 0;
// For all indices in s1.
for (int i=0; i<=n; i++) {
// For characters in range [a, z].
for (char ch='a'; ch<='z'; ch++) {
// THis is the updated string s1.
string updatedStr = s1.substr(0, i) + ch + s1.substr(i);
// Find lcs
int lcs2 = lcs(updatedStr, s2);
// If new lcs = original lcs + 1.
if (lcs2 == lcs1 + 1) ans++;
}
}
return ans;
}
int main() {
string s1 = "abab";
string s2 = "abc";
cout << waysToIncreaseLCSBy1(s1, s2);
return 0;
}
// Java program to get number of ways to increase
// LCS by 1
class GfG {
// Returns length of LCS for s1[0..m-1], s2[0..n-1]
static int lcs(String s1, String s2) {
int m = s1.length();
int n = s2.length();
// Initializing a matrix of size (m+1)*(n+1)
int[][] dp = new int[m + 1][n + 1];
// Building dp[m+1][n+1] in bottom-up fashion
for (int i = 1; i <= m; ++i) {
for (int j = 1; j <= n; ++j) {
if (s1.charAt(i - 1) == s2.charAt(j - 1))
dp[i][j] = dp[i - 1][j - 1] + 1;
else
dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
}
}
// dp[m][n] contains length of LCS for s1[0..m-1]
// and s2[0..n-1]
return dp[m][n];
}
// Method returns total ways to increase LCS length by 1
static int waysToIncreaseLCSBy1(String s1, String s2) {
int n = s1.length(), m = s2.length();
// Find the original LCS.
int lcs1 = lcs(s1, s2);
int ans = 0;
// For all indices in s1.
for (int i = 0; i <= n; i++) {
// For characters in range [a, z].
for (char ch = 'a'; ch <= 'z'; ch++) {
// This is the updated string s1.
String updatedStr = s1.substring(0, i) + ch + s1.substring(i);
// Find lcs
int lcs2 = lcs(updatedStr, s2);
// If new lcs = original lcs + 1.
if (lcs2 == lcs1 + 1) ans++;
}
}
return ans;
}
public static void main(String[] args) {
String s1 = "abab";
String s2 = "abc";
System.out.println(waysToIncreaseLCSBy1(s1, s2));
}
}
# Python program to get number of ways to increase
# LCS by 1
# Returns length of LCS for s1[0..m-1], s2[0..n-1]
def lcs(s1, s2):
m, n = len(s1), len(s2)
# Initializing a matrix of size (m+1)*(n+1)
dp = [[0] * (n + 1) for _ in range(m + 1)]
# Building dp[m+1][n+1] in bottom-up fashion
for i in range(1, m + 1):
for j in range(1, n + 1):
if s1[i - 1] == s2[j - 1]:
dp[i][j] = dp[i - 1][j - 1] + 1
else:
dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])
# dp[m][n] contains length of LCS for s1[0..m-1]
# and s2[0..n-1]
return dp[m][n]
# Method returns total ways to increase LCS length by 1
def waysToIncreaseLCSBy1(s1, s2):
n, m = len(s1), len(s2)
# Find the original LCS.
lcs1 = lcs(s1, s2)
ans = 0
# For all indices in s1.
for i in range(n + 1):
# For characters in range [a, z].
for ch in range(ord('a'), ord('z') + 1):
# This is the updated string s1.
updatedStr = s1[:i] + chr(ch) + s1[i:]
# Find lcs
lcs2 = lcs(updatedStr, s2)
# If new lcs = original lcs + 1.
if lcs2 == lcs1 + 1:
ans += 1
return ans
if __name__ == "__main__":
s1 = "abab"
s2 = "abc"
print(waysToIncreaseLCSBy1(s1, s2))
// C# program to get number of ways to increase
// LCS by 1
using System;
class GfG {
// Returns length of LCS for s1[0..m-1], s2[0..n-1]
static int lcs(string s1, string s2) {
int m = s1.Length;
int n = s2.Length;
// Initializing a matrix of size (m+1)*(n+1)
int[,] dp = new int[m + 1, n + 1];
// Building dp[m+1][n+1] in bottom-up fashion
for (int i = 1; i <= m; ++i) {
for (int j = 1; j <= n; ++j) {
if (s1[i - 1] == s2[j - 1])
dp[i, j] = dp[i - 1, j - 1] + 1;
else
dp[i, j] = Math.Max(dp[i - 1, j], dp[i, j - 1]);
}
}
// dp[m][n] contains length of LCS for s1[0..m-1]
// and s2[0..n-1]
return dp[m, n];
}
// Method returns total ways to increase LCS length by 1
static int waysToIncreaseLCSBy1(string s1, string s2) {
int n = s1.Length, m = s2.Length;
// Find the original LCS.
int lcs1 = lcs(s1, s2);
int ans = 0;
// For all indices in s1.
for (int i = 0; i <= n; i++) {
// For characters in range [a, z].
for (char ch = 'a'; ch <= 'z'; ch++) {
// This is the updated string s1.
string updatedStr = s1.Substring(0, i) + ch + s1.Substring(i);
// Find lcs
int lcs2 = lcs(updatedStr, s2);
// If new lcs = original lcs + 1.
if (lcs2 == lcs1 + 1) ans++;
}
}
return ans;
}
static void Main() {
string s1 = "abab";
string s2 = "abc";
Console.WriteLine(waysToIncreaseLCSBy1(s1, s2));
}
}
// JavaScript program to get number of ways to increase
// LCS by 1
// Returns length of LCS for s1[0..m-1], s2[0..n-1]
function lcs(s1, s2) {
let m = s1.length;
let n = s2.length;
// Initializing a matrix of size (m+1)*(n+1)
let dp = Array.from({ length: m + 1 }, () => Array(n + 1).fill(0));
// Building dp[m+1][n+1] in bottom-up fashion
for (let i = 1; i <= m; i++) {
for (let j = 1; j <= n; j++) {
if (s1[i - 1] === s2[j - 1])
dp[i][j] = dp[i - 1][j - 1] + 1;
else
dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
}
}
// dp[m][n] contains length of LCS for s1[0..m-1]
// and s2[0..n-1]
return dp[m][n];
}
// Method returns total ways to increase LCS length by 1
function waysToIncreaseLCSBy1(s1, s2) {
let n = s1.length, m = s2.length;
// Find the original LCS.
let lcs1 = lcs(s1, s2);
let ans = 0;
// For all indices in s1.
for (let i = 0; i <= n; i++) {
// For characters in range [a, z].
for (let ch = 97; ch <= 122; ch++) {
// This is the updated string s1.
let updatedStr = s1.substring(0, i) +
String.fromCharCode(ch) + s1.substring(i);
// Find lcs
let lcs2 = lcs(updatedStr, s2);
// If new lcs = original lcs + 1.
if (lcs2 === lcs1 + 1) ans++;
}
}
return ans;
}
let s1 = "abab";
let s2 = "abc";
console.log(waysToIncreaseLCSBy1(s1, s2));
Time Complexity: O(n * 26 * (n + n*m)) which can be simplified into O(n^2 * m). For each index in the first string, we can add 26 characters. Concatenating substrings takes O(n) time and finding LCS take O(n*m) time.
Auxiliary Space: O(n * m) due to memo matrix.
[Efficient Approach] Using Prefix and Suffix DP
The idea is to use prefix and suffix dynamic programming (DP) tables to precompute the LCS lengths for all possible prefixes and suffixes of S1 and S2. By inserting a character into S1 at every possible position and checking if the combined LCS length from the prefix and suffix tables increases the overall LCS length by one, we can count the number of valid insertions.
In the previous approach, LCS of updated string S1 and string S2 was repeatedly calculated for all possible characters and indices. But in this approach, the LCS for all prefixes and suffixes of S1 and S2 are stored, therefore we can find the LCS of any prefix or suffix of S1 and S2 in O(1) time.
If by adding a character after i 'th index in string S1 and matching this character with the character at j 'th index in string S2, we will find the LCS of prefix of S1 ( from index 0 to i) and S2 (from index 0 to j-1) and LCS of suffix of S1 (from index i + 1) and S2 (from j + 1). If the sum of prefix and suffix LCS equals LCS of complete strings, it means LCS can be increased ( by including the matched character added after i 'th index).
The LCS length increases by one if:
lcsl[i][j-1] + lcsr[i+1][j+1] == lcsl[n][m], where:lcsl[i][j-1] is the LCS length of the prefix S1[0..i-1] and S2[0..j-2].lcsr[i+1][j+1] is the LCS length of the suffix S1[i..n-1] and S2[j..m-1].lcsl[n][m] is the LCS length of the entire strings S1 and S2.
Step by step approach:
- Compute the prefix LCS table (
lcsl) for all prefixes of S1 and S2. - Compute the suffix LCS table (
lcsr) for all suffixes of S1 and S2. - Iterate over all possible insertion positions in
S1. - For each insertion position, try inserting every lowercase character (
a to z). - For each character, check if inserting it at the current position increases the LCS length by one using the prefix and suffix LCS tables. If the condition is satisfied, increment the answer.
- Return the total count of valid insertions.
Note: The break statement is important because it ensures that we count each valid insertion only once for a given position i in S1. Without the break, we would overcount by considering multiple valid j positions in S2 for the same character c at position i, which would incorrectly increment the result. The goal is to count the number of ways (characters or indices) in S1, not different combinations of i and j.
C++
// C++ program to get number of ways to increase
// LCS by 1
#include <bits/stdc++.h>
using namespace std;
// Method returns total ways to increase LCS length by 1
int waysToIncreaseLCSBy1(string s1, string s2) {
int n = s1.length(), m = s2.length();
vector<vector<int>> lcsl(n+2, vector<int>(m+2, 0));
vector<vector<int>> lcsr(n+2, vector<int>(m+2, 0));
// Filling LCS array for prefix substrings
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
if (s1[i-1] == s2[j-1])
lcsl[i][j] = 1 + lcsl[i-1][j-1];
else
lcsl[i][j] = max(lcsl[i-1][j],
lcsl[i][j-1]);
}
}
// Filling LCS array for suffix substrings
for (int i = n; i >= 1; i--) {
for (int j = m; j >= 1; j--) {
if (s1[i-1] == s2[j-1])
lcsr[i][j] = 1 + lcsr[i+1][j+1];
else
lcsr[i][j] = max(lcsr[i+1][j],
lcsr[i][j+1]);
}
}
// Looping for all possible insertion positions
// in first string
int ans = 0;
for (int i=0; i<=n; i++) {
// Trying all possible lower case characters
for (char c='a'; c<='z'; c++) {
// Find all positions of char c
// in string s2.
for (int j=1; j<=m; j++) {
if (s2[j-1] == c) {
// If adding char c at position (i+1) increases
// LCS, increment ans and break.
if (lcsl[i][j-1] + lcsr[i+1][j+1] == lcsl[n][m]) {
ans++;
break;
}
}
}
}
}
return ans;
}
int main() {
string s1 = "abab";
string s2 = "abc";
cout << waysToIncreaseLCSBy1(s1, s2);
return 0;
}
// Java program to get number of ways to increase
// LCS by 1
import java.util.*;
class GfG {
// Method returns total ways to increase LCS length by 1
static int waysToIncreaseLCSBy1(String s1, String s2) {
int n = s1.length(), m = s2.length();
int[][] lcsl = new int[n+2][m+2];
int[][] lcsr = new int[n+2][m+2];
// Filling LCS array for prefix substrings
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
if (s1.charAt(i-1) == s2.charAt(j-1))
lcsl[i][j] = 1 + lcsl[i-1][j-1];
else
lcsl[i][j] = Math.max(lcsl[i-1][j], lcsl[i][j-1]);
}
}
// Filling LCS array for suffix substrings
for (int i = n; i >= 1; i--) {
for (int j = m; j >= 1; j--) {
if (s1.charAt(i-1) == s2.charAt(j-1))
lcsr[i][j] = 1 + lcsr[i+1][j+1];
else
lcsr[i][j] = Math.max(lcsr[i+1][j], lcsr[i][j+1]);
}
}
// Looping for all possible insertion positions
// in first string
int ans = 0;
for (int i = 0; i <= n; i++) {
// Trying all possible lower case characters
for (char c = 'a'; c <= 'z'; c++) {
// Find all positions of char c
// in string s2.
for (int j = 1; j <= m; j++) {
if (s2.charAt(j-1) == c) {
// If adding char c at position (i+1) increases
// LCS, increment ans and break.
if (lcsl[i][j-1] + lcsr[i+1][j+1] == lcsl[n][m]) {
ans++;
break;
}
}
}
}
}
return ans;
}
public static void main(String[] args) {
String s1 = "abab";
String s2 = "abc";
System.out.println(waysToIncreaseLCSBy1(s1, s2));
}
}
# Python program to get number of ways to increase
# LCS by 1
# Method returns total ways to increase LCS length by 1
def waysToIncreaseLCSBy1(s1, s2):
n, m = len(s1), len(s2)
lcsl = [[0] * (m + 2) for _ in range(n + 2)]
lcsr = [[0] * (m + 2) for _ in range(n + 2)]
# Filling LCS array for prefix substrings
for i in range(1, n + 1):
for j in range(1, m + 1):
if s1[i - 1] == s2[j - 1]:
lcsl[i][j] = 1 + lcsl[i - 1][j - 1]
else:
lcsl[i][j] = max(lcsl[i - 1][j], lcsl[i][j - 1])
# Filling LCS array for suffix substrings
for i in range(n, 0, -1):
for j in range(m, 0, -1):
if s1[i - 1] == s2[j - 1]:
lcsr[i][j] = 1 + lcsr[i + 1][j + 1]
else:
lcsr[i][j] = max(lcsr[i + 1][j], lcsr[i][j + 1])
# Looping for all possible insertion positions
# in first string
ans = 0
for i in range(n + 1):
# Trying all possible lower case characters
for c in range(ord('a'), ord('z') + 1):
# Find all positions of char c
# in string s2.
for j in range(1, m + 1):
if s2[j - 1] == chr(c):
# If adding char c at position (i+1) increases
# LCS, increment ans and break.
if lcsl[i][j - 1] + lcsr[i + 1][j + 1] == lcsl[n][m]:
ans += 1
break
return ans
if __name__ == "__main__":
s1 = "abab"
s2 = "abc"
print(waysToIncreaseLCSBy1(s1, s2))
// C# program to get number of ways to increase
// LCS by 1
using System;
class GfG {
// Method returns total ways to increase LCS length by 1
static int waysToIncreaseLCSBy1(string s1, string s2) {
int n = s1.Length, m = s2.Length;
int[,] lcsl = new int[n+2, m+2];
int[,] lcsr = new int[n+2, m+2];
// Filling LCS array for prefix substrings
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
if (s1[i-1] == s2[j-1])
lcsl[i, j] = 1 + lcsl[i-1, j-1];
else
lcsl[i, j] = Math.Max(lcsl[i-1, j], lcsl[i, j-1]);
}
}
// Filling LCS array for suffix substrings
for (int i = n; i >= 1; i--) {
for (int j = m; j >= 1; j--) {
if (s1[i-1] == s2[j-1])
lcsr[i, j] = 1 + lcsr[i+1, j+1];
else
lcsr[i, j] = Math.Max(lcsr[i+1, j], lcsr[i, j+1]);
}
}
// Looping for all possible insertion positions
// in first string
int ans = 0;
for (int i = 0; i <= n; i++) {
// Trying all possible lower case characters
for (char c = 'a'; c <= 'z'; c++) {
// Find all positions of char c
// in string s2.
for (int j = 1; j <= m; j++) {
if (s2[j-1] == c) {
// If adding char c at position (i+1) increases
// LCS, increment ans and break.
if (lcsl[i, j-1] + lcsr[i+1, j+1] == lcsl[n, m]) {
ans++;
break;
}
}
}
}
}
return ans;
}
static void Main() {
string s1 = "abab";
string s2 = "abc";
Console.WriteLine(waysToIncreaseLCSBy1(s1, s2));
}
}
// JavaScript program to get number of ways to increase
// LCS by 1
// Method returns total ways to increase LCS length by 1
function waysToIncreaseLCSBy1(s1, s2) {
let n = s1.length, m = s2.length;
let lcsl = Array.from({ length: n + 2 }, () => Array(m + 2).fill(0));
let lcsr = Array.from({ length: n + 2 }, () => Array(m + 2).fill(0));
// Filling LCS array for prefix substrings
for (let i = 1; i <= n; i++) {
for (let j = 1; j <= m; j++) {
if (s1[i - 1] === s2[j - 1])
lcsl[i][j] = 1 + lcsl[i - 1][j - 1];
else
lcsl[i][j] = Math.max(lcsl[i - 1][j], lcsl[i][j - 1]);
}
}
// Filling LCS array for suffix substrings
for (let i = n; i >= 1; i--) {
for (let j = m; j >= 1; j--) {
if (s1[i - 1] === s2[j - 1])
lcsr[i][j] = 1 + lcsr[i + 1][j + 1];
else
lcsr[i][j] = Math.max(lcsr[i + 1][j], lcsr[i][j + 1]);
}
}
// Looping for all possible insertion positions
// in first string
let ans = 0;
for (let i = 0; i <= n; i++) {
// Trying all possible lower case characters
for (let c = 'a'.charCodeAt(0); c <= 'z'.charCodeAt(0); c++) {
let charC = String.fromCharCode(c);
// Find all positions of char c
// in string s2.
for (let j = 1; j <= m; j++) {
if (s2[j - 1] === charC) {
// If adding char c at position (i+1) increases
// LCS, increment ans and break.
if (lcsl[i][j - 1] + lcsr[i + 1][j + 1] === lcsl[n][m]) {
ans++;
break;
}
}
}
}
}
return ans;
}
let s1 = "abab";
let s2 = "abc";
console.log(waysToIncreaseLCSBy1(s1, s2));
Time Complexity : O(n * m), where n is the length of string s1 and m is the length of string s2.
Auxiliary Space : O(n * m)
Similar Reads
Basics & Prerequisites
Data Structures
Array Data StructureIn this article, we introduce array, implementation in different popular languages, its basic operations and commonly seen problems / interview questions. An array stores items (in case of C/C++ and Java Primitive Arrays) or their references (in case of Python, JS, Java Non-Primitive) at contiguous
3 min read
String in Data StructureA string is a sequence of characters. The following facts make string an interesting data structure.Small set of elements. Unlike normal array, strings typically have smaller set of items. For example, lowercase English alphabet has only 26 characters. ASCII has only 256 characters.Strings are immut
2 min read
Hashing in Data StructureHashing is a technique used in data structures that efficiently stores and retrieves data in a way that allows for quick access. Hashing involves mapping data to a specific index in a hash table (an array of items) using a hash function. It enables fast retrieval of information based on its key. The
2 min read
Linked List Data StructureA linked list is a fundamental data structure in computer science. It mainly allows efficient insertion and deletion operations compared to arrays. Like arrays, it is also used to implement other data structures like stack, queue and deque. Here’s the comparison of Linked List vs Arrays Linked List:
2 min read
Stack Data StructureA Stack is a linear data structure that follows a particular order in which the operations are performed. The order may be LIFO(Last In First Out) or FILO(First In Last Out). LIFO implies that the element that is inserted last, comes out first and FILO implies that the element that is inserted first
2 min read
Queue Data StructureA Queue Data Structure is a fundamental concept in computer science used for storing and managing data in a specific order. It follows the principle of "First in, First out" (FIFO), where the first element added to the queue is the first one to be removed. It is used as a buffer in computer systems
2 min read
Tree Data StructureTree Data Structure is a non-linear data structure in which a collection of elements known as nodes are connected to each other via edges such that there exists exactly one path between any two nodes. Types of TreeBinary Tree : Every node has at most two childrenTernary Tree : Every node has at most
4 min read
Graph Data StructureGraph Data Structure is a collection of nodes connected by edges. It's used to represent relationships between different entities. If you are looking for topic-wise list of problems on different topics like DFS, BFS, Topological Sort, Shortest Path, etc., please refer to Graph Algorithms. Basics of
3 min read
Trie Data StructureThe Trie data structure is a tree-like structure used for storing a dynamic set of strings. It allows for efficient retrieval and storage of keys, making it highly effective in handling large datasets. Trie supports operations such as insertion, search, deletion of keys, and prefix searches. In this
15+ min read
Algorithms
Searching AlgorithmsSearching algorithms are essential tools in computer science used to locate specific items within a collection of data. In this tutorial, we are mainly going to focus upon searching in an array. When we search an item in an array, there are two most common algorithms used based on the type of input
2 min read
Sorting AlgorithmsA Sorting Algorithm is used to rearrange a given array or list of elements in an order. For example, a given array [10, 20, 5, 2] becomes [2, 5, 10, 20] after sorting in increasing order and becomes [20, 10, 5, 2] after sorting in decreasing order. There exist different sorting algorithms for differ
3 min read
Introduction to RecursionThe process in which a function calls itself directly or indirectly is called recursion and the corresponding function is called a recursive function. A recursive algorithm takes one step toward solution and then recursively call itself to further move. The algorithm stops once we reach the solution
14 min read
Greedy AlgorithmsGreedy algorithms are a class of algorithms that make locally optimal choices at each step with the hope of finding a global optimum solution. At every step of the algorithm, we make a choice that looks the best at the moment. To make the choice, we sometimes sort the array so that we can always get
3 min read
Graph AlgorithmsGraph is a non-linear data structure like tree data structure. The limitation of tree is, it can only represent hierarchical data. For situations where nodes or vertices are randomly connected with each other other, we use Graph. Example situations where we use graph data structure are, a social net
3 min read
Dynamic Programming or DPDynamic Programming is an algorithmic technique with the following properties.It is mainly an optimization over plain recursion. Wherever we see a recursive solution that has repeated calls for the same inputs, we can optimize it using Dynamic Programming. The idea is to simply store the results of
3 min read
Bitwise AlgorithmsBitwise algorithms in Data Structures and Algorithms (DSA) involve manipulating individual bits of binary representations of numbers to perform operations efficiently. These algorithms utilize bitwise operators like AND, OR, XOR, NOT, Left Shift, and Right Shift.BasicsIntroduction to Bitwise Algorit
4 min read
Advanced
Segment TreeSegment Tree is a data structure that allows efficient querying and updating of intervals or segments of an array. It is particularly useful for problems involving range queries, such as finding the sum, minimum, maximum, or any other operation over a specific range of elements in an array. The tree
3 min read
Pattern SearchingPattern searching algorithms are essential tools in computer science and data processing. These algorithms are designed to efficiently find a particular pattern within a larger set of data. Patten SearchingImportant Pattern Searching Algorithms:Naive String Matching : A Simple Algorithm that works i
2 min read
GeometryGeometry is a branch of mathematics that studies the properties, measurements, and relationships of points, lines, angles, surfaces, and solids. From basic lines and angles to complex structures, it helps us understand the world around us.Geometry for Students and BeginnersThis section covers key br
2 min read
Interview Preparation
Practice Problem